The given expression is:

$\frac{(x + 5)(4x^2 - 9)}{(x^2 + 6x + 5)(2x^2 + 3x - 9)}$

Step 1: Factorizing the components

Let's first factor each part of the expression:

Numerator:

1. $x + 5$ is already factored.
2. $4x^2 - 9$ is a difference of squares: $4x^2 - 9 = (2x - 3)(2x + 3)$

Denominator:

3. $x^2 + 6x + 5$ factors as: $x^2 + 6x + 5 = (x + 5)(x + 1)$
4. $2x^2 + 3x - 9$ factors as: $2x^2 + 3x - 9 = (2x - 3)(x + 3)$

Step 2: Substitute factored terms into the expression

Now, we rewrite the entire expression with the factored terms: $\frac{(x + 5)(2x - 3)(2x + 3)}{(x + 5)(x + 1)(2x - 3)(x + 3)}$

Step 3: Simplify by canceling common factors

The common factors in the numerator and denominator are $(x + 5), (2x - 3),$ and $(x + 3)$. Canceling these gives: $\frac{2x + 3}{x + 1}$

Final Answer:

The simplified expression is: $\frac{2x + 3}{x + 1}$

Let me know if you'd like more details on the factorization or the simplification steps.

Follow-up Questions:

1. How would you factor a trinomial like $x^2 + 6x + 5$?
2. Can you explain why $4x^2 - 9$ is a difference of squares?
3. How do you determine which terms can be canceled when simplifying fractions?
4. What happens to the expression if $x = -1$? Why is this important?
5. Can all quadratics be factored this way?

Tip: Always check for common factors first when simplifying complex fractions—it often makes the problem easier to solve!